Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const long long int INF = 1e12;
const int N = 1e5 + 1;
vector<int> adj[N], depth(N, 0);
vector<int> mins[17][N], persons[N];
int dp[17][N];
vector<int> merge(vector<int> &a, vector<int> &b, int limit) {
  vector<int> res;
  int i = 0, j = 0;
  while (i < a.size() && j < b.size()) {
    if (a[i] < b[j])
      res.push_back(a[i]), i++;
    else
      res.push_back(b[j]), j++;
  }
  while (i < a.size()) res.push_back(a[i]), i++;
  while (j < b.size()) res.push_back(b[j]), j++;
  while ((int)res.size() > limit) res.pop_back();
  return res;
}
void dfs(int v, int parent, int l) {
  depth[v] = l;
  dp[0][v] = parent;
  if (parent != -1) mins[0][v] = persons[parent];
  for (int u : adj[v]) {
    if (u == parent) continue;
    dfs(u, v, l + 1);
  }
}
int LCA(int u, int v, int n) {
  if (depth[u] > depth[v]) swap(u, v);
  int dist = depth[v] - depth[u];
  while (dist > 0) {
    int i = log2(dist);
    v = dp[i][v];
    dist -= (1 << i);
  }
  if (u == v) return v;
  for (int i = log2(n); i >= 0; i--) {
    if (dp[i][v] != -1 && dp[i][u] != dp[i][v]) u = dp[i][u], v = dp[i][v];
  }
  return dp[0][v];
}
void compute(vector<int> &res, int u, int v, int limit) {
  int dist = depth[v] - depth[u];
  while (dist > 0) {
    int i = log2(dist);
    res = merge(res, mins[i][v], limit);
    v = dp[i][v];
    dist -= (1 << i);
  }
}
int parent(int v, int dist) {
  while (dist > 0) {
    int i = log2(dist);
    v = dp[i][v];
    dist -= (1 << i);
  }
  return v;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m, q;
  cin >> n >> m >> q;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v), adj[v].push_back(u);
  }
  for (int i = 1; i <= m; i++) {
    int c;
    cin >> c;
    if (persons[c].size() < 10) persons[c].push_back(i);
  }
  dfs(1, -1, 0);
  for (int i = 1; i <= log2(n); i++) {
    for (int j = 1; j <= n; j++) {
      if (dp[i - 1][j] != -1) {
        dp[i][j] = dp[i - 1][dp[i - 1][j]];
        mins[i][j] = merge(mins[i - 1][j], mins[i - 1][dp[i - 1][j]], 10);
      }
    }
  }
  while (q--) {
    int u, v, a;
    cin >> u >> v >> a;
    vector<int> res;
    int lca = LCA(u, v, n);
    if (lca == u) {
      res = persons[v];
      compute(res, lca, v, a);
    } else if (lca == v) {
      res = persons[u];
      compute(res, lca, u, a);
    } else {
      int child = parent(v, depth[v] - depth[lca] - 1);
      res = persons[u];
      res = merge(res, persons[v], 10);
      compute(res, lca, u, a);
      compute(res, child, v, a);
    }
    while ((int)res.size() > a) res.pop_back();
    cout << res.size() << " ";
    for (int i : res) cout << i << " ";
    cout << "\n";
  }
  return 0;
}